Let the function be #f(x)#:
#f'(x) = lim_(h ->0) (f(x + h) - f(x))/h#
#f'(x) = lim_(h->0) (1/((x + h)^2 - 1) - 1/(x^2 - 1))/h#
#f'(x) = lim_(h->0) (1/(x^2 + 2xh + h^2 - 1) - 1/(x^2 - 1))/h#
#f'(x) = lim_(h->0) ((x^2 - 1 - (x^2 + 2xh + h^2 - 1))/((x^2 - 1)(x^2 + 2xh + h^2 - 1)))/h#
#f'(x) = lim_(h->0) ((x^2 - 1-x^2 - 2xh-h^2+ 1)/((x^2 - 1)(x^2+ 2xh + h^2 - 1)))/h#
#f'(x) = lim_(h->0) (-2xh + h^2)/(h(x^4 + 2x^3h + x^2h^2 - x^2 - x^2-2xh-h^2 + 1))#
#f'(x) = lim_(h->0) (cancel(h)(-2x + h))/(cancel(h)(x^4 + 2x^3h + x^2h^2+ 2xh + h^2 + 1 - 2x^2)#
Substituting:
#f'(x) = (-2x + 0)/(x^4 + 2x^3(0) + x^2(0)^2 + 2x(0) + 0^2 + 1 - 2x^2)#
#f'(x) = -(2x)/(x^4 - 2x^2+ 1)#
Hence, the derivative of #f(x) = 1/(x^2 - 1)# is #f'(x) = -(2x)/(x^4 - 2x^2 + 1)#. All that we have left to do is plug in #x = 3# into our derivative:
#f'(3) = -(2 xx 3)/(3^4 - 2(3)^2 + 1)#
#f'(3) = -6/(81 - 18 + 1)#
#f'(3) = -6/(64)#
#f'(3) = -3/32#
In summary, when #f'(3)# is evaluated inside the derivative of #f(x) = 1/(x^2 - 1)#, a result of #-3/32# is obtained.
Hopefully this helps!
